改写 y = (x^n)/(1-x) = {[(x^n)-1]+1}/(1-x) = {[x^(n-1)] + [x^(n-2)] + … + x + 1} + 1/(1-x),因此, y' = {(n-1)[x^(n-2)] + (n-2)[x^(n-3)] + … + 1} + 1/(1-x)^2, y" = {(n-1)(n-2)[x^(n-3)] + ...
